When $(1+x)^y (1-x)^z$ is smaller than 1?

Question

I need to find when: $$(1+x)^y (1-x)^z$$ is smaller than 1, given $0<x<1$, $0<y<1$, and $0<z<1$

A sufficient condition is $z\geq y$, but I cannot find a necessary and sufficient condition.

Answer 1

Given the restrictions on $x$, $y$, and $z$ the given inequality is equivalent with $${z\over y}>{\log(1+x)\over\log{1\over 1-x}}\ .\tag{1}$$ Since ${1\over1-x}>1+x$ when $0<x<1$ the right hand side of $(1)$ is $<1$ when $0<x<1$. Therefore $z\geq y$ is sufficient, but not necessary for the given inequality to hold. Plotting the right hand side of $(1)$ as a function of $x$ one sees that, e.g., the condition $${z\over y}\geq1-0.793 x$$ is also sufficient.

Answer 2

Your sufficient condition $(z\ge y)$ is also necessary for the inequality to hold for all $x$ in the interval. If $z<y$, we set $y=z+t$, for $t\in(0,1)$. Raising both sides to $1/z$ power, we want to prove that there is some $x\in(0,1)$ with $$(1+x)^{1+t/z}(1-x)\ge 1$$

Considering $f(x)=(1+x)^{1+t/z}(1-x)$ (which is differentiable and continuous), we see that $f(0)=1$, and also $$f'(x)=-(x+1)^{t/z}((1+t/z)(x-1)+x+1)$$ Hence $f'(0)=t/z>0$. So $f(\epsilon)>1$ for some $\epsilon>0$.